Respuesta :
Answer: The volume of barium hydroxide is 183 mL.
Explanation:
To calculate the moles of cadmium nitrate, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex] .....(1)
Molarity of [tex]MnSO_4[/tex] = 0.796 M
Volume of [tex]MnSO_4[/tex] = 161 mL = 0.161 L (Conversion factor: 1 L = 1000 mL)
Putting values in equation 1, we get:
[tex]0.796mol/L=\frac{\text{Moles of }MnSO_4}{0.161L}\\\\\text{Moles of }MnSO_4=0.13mol[/tex]
The chemical equation for the reaction of manganese sulfate and barium hydroxide follows:
[tex]MnSO_4(aq.)+Ba(OH)_2(aq.)\rightarrow Mn(OH)_2(s)+BaSO_4(aq.)[/tex]
By Stoichiometry of the reaction:
1 mole of manganese sulfate reacts with 1 mole of barium hydroxide.
So, 0.13 moles of manganese sulfate will react with = [tex]\frac{1}{1}\times 0.13=0.13mol[/tex] of barium hydroxide
Now, calculating the volume of barium hydroxide by using equation 1, we get:
Moles of barium hydroxide = 0.13 moles
Molarity of barium hydroxide = 0.710 M
Putting values in equation 1, we get:
[tex]0.710mol/L=\frac{0.13mol}{\text{Volume of barium hydroxide}}\\\\\text{Volume of barium hydroxide}=0.183L[/tex]
Converting this into milliliters, we use the conversion factor:
1 L = 1000 mL
So, [tex]0.183L=0.183\times 1000=183mL[/tex]
Hence, the volume of barium hydroxide is 183 mL.
